Matrix X and Y Solutions for Equations XA=B and AY=B

[Physicsissuef]

Homework Statement

I have two 3x3 matrices A and B

and my problem is to find XA=B and AY=B

Isn't XA=B and AY=B

Homework Equations

The Attempt at a Solution

$$X=A^-^1 B$$

$$Y=A^-^1 B$$

or I am wrong?

 

[malawi_glenn]

remember that matrix multiplication is non-commutative

so yes you are wrong on the $$X=A^-^1 B$$, if XA=B

 

[Physicsissuef]

remember that matrix multiplication is non-commutative

so yes you are wrong on the $$X=A^-^1 B$$, if XA=B

Can you tell me please, what is the correct formula?

 

[D H]

You can pre- or post-multiply a matrix equality by a matrix to yield another matrix equality: If A=B then CA = CB and AD = BD for all matrices C, D of the correct dimensionality. Use this plus the fact that matrix multiplication is associative to find a form that eliminates the matrix A from XA=B.

 

[Physicsissuef]

You can pre- or post-multiply a matrix equality by a matrix to yield another matrix equality: If A=B then CA = CB and AD = BD for all matrices C, D of the correct dimensionality. Use this plus the fact that matrix multiplication is associative to find a form that eliminates the matrix A from XA=B.

Sorry, but I don't understand what you want to say... Can you please write what equals X, and what equals Y?

$$X= ?? Y=??$$

 

[e(ho0n3]

Why don't you write out XA = AY, compare the entries and figure out what X and Y could be.

 

[malawi_glenn]

We don't give answers here for free!

You must multiply matrices from the correct order!

Take this for example:

XA=B

How would you do to eliminate X ? By multiplying inv(A) from the left?:
Then you'll get:

inv(A)XA = inv(A)B

And that is NOT you want right? So how would you do it?

 

[K.J.Healey]

You can multiply from both sides in matrix:
If you have
A=B
Then you can have
CA = CB
OR
AC = BC

The C can come in on the right or left. But you have to keep track.

 

[Physicsissuef]

So X=inv(a)B
and inv(a)AY=inv(a)B , so Y=inv(a)B

hm...

 

[malawi_glenn]

Well the Y is correct, but not the X, try again

you have:

XA=B

 

[Physicsissuef]

XA=B

inv(x)XA=inv(x)B

A=inv(x)B ??

 

[malawi_glenn]

well yes, but you want to solve for X right?

Read post #8, by K.J.Healey (T, 19:38)

 

[D H]

You are supposed to be solving for X, not A. Why are you insisting on multiplying on the left? XA has two sides (left and right), and so does B.

 

[Physicsissuef]

XA=B

XA(inv (a))=B(inv (a))

X=B(inv(a))

like this?

 

[malawi_glenn]

yes! Good job :-)

 

[Physicsissuef]

Thank you very much, guys. I love youuuu...

 

[orion216]

Hi all,

Pardon me for digging up this old thread, but it is related to what I am asking.
Is there some matrix manipulation that transforms equation of type xA = B into A'x' = B'? in other words, transform Unknown.KNOWN = KNOWN into KNOWN.Unknown = KNOWN.
Thanks a bunch!